A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it woul have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10km/h. The distance between the two cities is

Options

A. 540km
B. 420km
C. 600km
D. 620km

Answer

To solve the problem step by step, let's denote the original speed of the car as \( S \) km/h and the distance between the two cities \( P \) and \( Q \) as \( D \) km.

### Step 1: Set up the equations based on the problem statement.

1. The time taken to travel the distance at speed \( S \) is given by:
\[ \text{Time} = \frac{D}{S} \]

2. If the speed is increased by 10 km/h, the new speed becomes \( S + 10 \) km/h. The time taken at this speed is:
\[ \text{New Time} = \frac{D}{S + 10} \]

3. According to the problem, this new time is 1 hour less than the original time:
\[ \frac{D}{S} - \frac{D}{S + 10} = 1 \quad \text{(Equation 1)} \]

### Step 2: Set up the second equation based on the further increase in speed.

1. If the speed is increased by another 10 km/h (making it \( S + 20 \) km/h), the time taken at this speed is:
\[ \text{Newer Time} = \frac{D}{S + 20} \]

2. This time is 45 minutes (or \( \frac{3}{4} \) hours) less than the time taken at speed \( S + 10 \):
\[ \frac{D}{S + 10} - \frac{D}{S + 20} = \frac{3}{4} \quad \text{(Equation 2)} \]

### Step 3: Solve Equation 1.

From Equation 1:
\[ \frac{D}{S} - \frac{D}{S + 10} = 1 \]
Multiply through by \( S(S + 10) \):
\[ D(S + 10) - DS = S(S + 10) \]
\[ 10D = S(S + 10) \]
\[ D = \frac{S(S + 10)}{10} \quad \text{(Equation 3)} \]

### Step 4: Solve Equation 2.

From Equation 2:
\[ \frac{D}{S + 10} - \frac{D}{S + 20} = \frac{3}{4} \]
Multiply through by \( (S + 10)(S + 20) \):
\[ D(S + 20) - D(S + 10) = \frac{3}{4}(S + 10)(S + 20) \]
\[ 10D = \frac{3}{4}(S + 10)(S + 20) \]
Substituting Equation 3 into this gives:
\[ 10 \left(\frac{S(S + 10)}{10}\right) = \frac{3}{4}(S + 10)(S + 20) \]
\[ S(S + 10) = \frac{3}{4}(S^2 + 30S + 200) \]

### Step 5: Clear the fraction and simplify.

Multiply through by 4:
\[ 4S(S + 10) = 3(S^2 + 30S + 200) \]
\[ 4S^2 + 40S = 3S^2 + 90S + 600 \]
Rearranging gives:
\[ S^2 - 50S - 600 = 0 \]

### Step 6: Solve the quadratic equation.

Using the quadratic formula \( S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ S = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 1 \cdot (-600)}}{2 \cdot 1} \]
\[ S = \frac{50 \pm \sqrt{2500 + 2400}}{2} \]
\[ S = \frac{50 \pm \sqrt{4900}}{2} \]
\[ S = \frac{50 \pm 70}{2} \]
Calculating the two possible values:
1. \( S = \frac{120}{2} = 60 \)
2. \( S = \frac{-20}{2} = -10 \) (not valid since speed cannot be negative)

Thus, \( S = 60 \) km/h.

### Step 7: Find the distance \( D \).

Substituting \( S \) back into Equation 3:
\[ D = \frac{60(60 + 10)}{10} = \frac{60 \times 70}{10} = 420 \text{ km} \]

### Final Answer:
The distance between the two cities \( P \) and \( Q \) is **420 km**.

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